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Juli2301 [7.4K]
3 years ago
15

Evaluate 18 - 5x when x = -7​

Mathematics
1 answer:
sukhopar [10]3 years ago
3 0

Answer:

i thinks its -53

Step-by-step explanation:

18-5x

5x-7= -35

-35-18

-53

You might be interested in
Caleb and Mike both score the same amount of points in a game. Caleb has fifteen less than the total number of points. Mike has
elena-s [515]
Answer:
129 points
Step-by-step explanation:
Let L and C represent the scores of Luke and Caleb, respectively.
L = 2C -15 . . . . . Luke scored 15 less than twice the number Caleb did
L +C = 201 . . . . . they scored 201 points altogether
Add twice the second equation to the first:
2(L +C) + (L) = 2(201) + (2C -15)
3L +2C = 387 +2C . . . . simplify
3L = 387 . . . . . . . . . . . . .subtract 2C
L = 387/3 = 129 . . . . . . .divide by 3
Luke scored 129 points.
3 0
3 years ago
A.)0.406<br> B.)0.435<br> C.)0.303<br> D.)0.154
Serggg [28]
I think it should be option d
3 0
3 years ago
The circumference of the hub cap of a tire is 83.96 centimeters. Find the area of this hub cap. Use 3.14 for . Use pencil and pa
Bess [88]

Answer:

Area = 561.2485 square centimeter

If the circumference of the hub cap would have been smaller, then its radius would have been smaller and subsequently its area would also have been less.

Step-by-step explanation:

As we know the circumference of a circle is 2\pi r

The radius of the hub cap needs to be devised to determine the area

83.96 = 2* 3.14*r

r = \frac{83.96}{2*3.14}

r = 13.37 centimeters

Area of the hub cap = \pi r^{2}

Substituting the devised value of r in the above equation, we get -

3.14 * 13.37^2\\

561.2485 square centimeter

If the circumference of the hub cap would have been smaller, then its radius would have been smaller and subsequently its area would also have been less.

5 0
3 years ago
Change the decimal into a %:0.7
Tomtit [17]
Simply multiply 0.7 by 100. 0.7 x 100= 70%. The answer is 70%
8 0
3 years ago
Read 2 more answers
Please please help me
Alex73 [517]

Answer:

x = 5.5

Step-by-step explanation:

Given 2 secants intersecting the circle from a point outside the circle then

The product of the external part and the entire part of one secant is equal to the product of the external part and the entire part of the other secant, that is

x(x + 14) = 6(6 + 12)

x² + 14x = 6 × 18 = 108 ( subtract 108 from both sides )

x² + 14x - 108 = 0 ← in standard form

with a = 1, b = 14 and c = - 108

Using the quadratic formula to solve for x

x = ( - b ± \sqrt{b^2-4ac} ) / 2a

  = ( - 14 ± \sqrt{14^2-(4(1)(-108)} ) / 2

  = ( - 14 ± \sqrt{196+432} ) / 2

  = ( - 14 ± \sqrt{628} ) / 2

  x = \frac{-14-\sqrt{628} }{2} or x = \frac{-14+\sqrt{628} }{2}

  x = - 19.5 or x = 5.5 ( to 1 dec. place )

However x > 0 ⇒ x = 5.5

6 0
3 years ago
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